@mapleatha  I don’t have Maple 13 to check everything, but the above method works for Maple>=2015. Try one more way. First evaluate your integral in the usual way, then create the text line by clicking the letter T on the Toolbar. Next, just copy the output of the previous calculation into this text line and manually remove the parenthesis. See the screenshot below:

@Carl Love  But apparently my proposal  ( 'a >= b'  in 2d math) is the simplest solution to the problem.

@mmcdara 

@nm 

1. I do not understand the meaning of the "solution" that Mathematica returns. Do you understand it?

2.Maple is simply not designed to solve differential equations with such a heap of unknown functions in one equation. If we denote it like  H(t)=F(x(t),y(t)) , then we get the correct solution:

dsolve(diff(H(t),t)=0, H(t));                        

                              H(t) = _C1

In other words, the solution is any function of the form  F(x(t),y(t))  identically equal to some constant.

And how would you solve these tasks  a)  and c)  manually? Answer these simple questions and I will show you how it can be automated in Maple.

You need to formulate your question more clearly. For example, I did not understand:

1. What partitions are we talking about, into 2 summands or to any number of summands starting from two?
2. Should all terms in the partition be prime numbers?
3. What is Q(n)  and  {phi}  means?
4. What is the specific feature of the numbers 63 and 161?

@Abdoulaye  I won’t be able to help you until you figure out what each line of your code does (after that, help is no longer needed). If you do not want to understand this, then you can just use my code.

@findoc  This is not difficult. We simply project this solution (the red dot) on the coordinate axes with black dashed lines:

restart; with(plots):
sys := [p+x+.6*y-15, p+.3*x+.2*y-10, p+.5*x+y-14]:
sol:=solve(sys, [x, y, p])[];
A:=implicitplot3d(sys, x = 0 .. 10, y = 0 .. 10, p = 0 .. 10, style=surface, color=["LightBlue","LightGreen","Yellow"]):
B:=pointplot3d(eval([x,y,p],sol), color=red, symbol=solidsphere, symbolsize=15):
C:=plottools:-line(eval([x,y,p],sol),eval([x,0,0],sol),color=black,linestyle=3),plottools:-line(eval([x,y,p],sol),eval([0,y,0],sol),color=black,linestyle=3),plottools:-line(eval([x,y,p],sol),eval([0,0,p],sol),color=black,linestyle=dash):
display(A,B,C, axes=normal, orientation=[-20,80], lightmodel=light4);

@mathkid99  Compare the last lines of my and your codes.

@Carl Love  Here is it:

Download permanouuuuuuuuuuuuuuuuuun_new.mw

@JAMET  To get the curve equation from equations  t  and  t1  in the form  F(x,y)=0  you just need to eliminate the parameter  m   from these equations:

restart;  
t := y-m*x-p/(2*m) = 0; 
t1 := y+x/m+(1/2)*p1*m = 0; 
Sol:=eliminate({t,t1}, m);
Eq:=op(Sol[2])=0;
plots:-implicitplot(eval(Eq,[p=1,p1=-2]), x=-2..2,y=-2..2,color=red, scaling=constrained, size=[200,600]); 

@vv  I don’t understand where the discontinuous points come from? The integrand is obviously continuous. Isn't its antiderivative function to be continuous?

For example, if we plot  F , we see a continuous curve:

F:=Int(sqrt(1+((k*Pi)/l*cos((Pi*x)/l))^2), x=0..X):
plot(eval(F,[k=2,l=2]), X=0..6);

@Mariusz Iwaniuk  Of course you're right. The integrand  sqrt(1 + k^2*Pi^2*cos(Pi*x/l)^2/l^2)  in the original integral is positive, therefore the definite integral  (if  b>a) is also positive. Maple seems to find this primitive function  F1 =  -csgn(sin(Pi*x/l))*Elliptic E(cos(Pi*x/l), I*k*Pi/l)*l/Pi  incorrectly.

@Carl Love  for this info. I fixed the procedure  P  code with this in mind.

@ActiveUser  Present here in text form code that does not work in Maple 12 (so that I can test it). The upper value for the variable  n  in the procedure  P  is not indicated , because it is unknown in advance and can be large enough if the numbers  a  and  b  are close.

PS. I slightly modified the procedure  P  code. Check again how it works.